Conjugate of Dirichlet Character

This just follows because \(z^{-1} = \overline{z}\) if \(|z| = 1\). That is, negating the argument of a complex number and flipping it about the real axis are the same geometric action.

Proof

Since \(\gcd(a, q) = 1\), we can consider \(\chi\) as a group character when restricted to \(\mathbb{Z}_q^\ast\) by identifying \(a\) with the corresponding reduced residue class. Then because characters of finite group map to roots of unity, we know that \(\chi(a)\) is a root of unity, and hence \(\chi(a)^n = 1 \implies |\chi(a)|^n = 1 \implies |\chi(a)| = 1\).

Thus writing \(\chi(a) = e^{i\theta}\) we can see that